Timeline for A question regarding a claim of V. I. Arnold
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 8, 2023 at 19:21 | comment | added | Z. M | @ToddTrimble I guess that he intended that one see that $(x^2+3x+2)^{-1}=(x+1)^{-1}-(x+2)^{-1}$. | |
| May 9, 2011 at 20:21 | comment | added | Todd Trimble | @BS: thanks. Yes, it's quite obvious now. | |
| May 9, 2011 at 20:19 | comment | added | Todd Trimble | Thank you for making those available, MathOMan! What an incredible batch of problems! But I must confess that I don't quite see the point of (for example) exercise no. 5. I can only imagine someone solving it as an American first-year calculus student would, only to have it angrily rejected by Arnold as being too... what? Formulaic? | |
| May 9, 2011 at 18:56 | history | edited | Andrey Rekalo | CC BY-SA 3.0 |
added 717 characters in body
|
| May 9, 2011 at 18:33 | comment | added | MathOMan | This is the second question of Arnold's Trivium. You can find the solutions to nearly all those questions (except for 27, 41, 51, 58, 68, 69, 70, 73, 74) here: mathoman.com/index.php/… | |
| May 9, 2011 at 18:12 | history | edited | Andrey Rekalo | CC BY-SA 3.0 |
added 1 characters in body
|
| May 9, 2011 at 18:08 | comment | added | Andrey Rekalo | @BS and Todd Trimble: Thank you for the comments. | |
| May 9, 2011 at 18:07 | history | undeleted | Andrey Rekalo | ||
| May 9, 2011 at 18:07 | history | edited | Andrey Rekalo | CC BY-SA 3.0 |
deleted 263 characters in body; added 126 characters in body
|
| May 9, 2011 at 17:59 | history | deleted | Andrey Rekalo | ||
| May 9, 2011 at 17:56 | history | edited | Andrey Rekalo | CC BY-SA 3.0 |
added 98 characters in body
|
| May 9, 2011 at 17:54 | comment | added | BS. | of course I meant $f_c^{-1}=c^{-1}f^{-1}$ | |
| May 9, 2011 at 17:53 | comment | added | BS. | @Todd and Andrey : the limit is $-(f'(0))^{k+1}$, when $f(x)-g(x)=ax^k+O(x^{k+1})$, $k>1$, $a\neq 0$. The proof is to observe that replacing $f$, $g$ by $f_c(x)=f(cx)$, $g_c(x)=g(cx)$, the quotient for $f_c,g_c$ is then equivalent (at $0$) to $c^{k+1}$ times that for $f,g$ (note that $f_c^{-1}=c^{-1}f$). One is now reduced to the case $f'(0)=g'(0)=1$. | |
| May 9, 2011 at 17:01 | comment | added | Todd Trimble | Is the generalization supposed to be that that limit quotient is $-(f'(0))^{2n+2}$ if $n$ is the least integer where $f^{2n+1}(0)$ and $g^{2n+1}(0)$ differ? | |
| May 9, 2011 at 16:51 | comment | added | BS. | Well, isn't your condition that $f$ and $g$ are different odd analytic functions ? And sure enough, $\sin$ and $\tan$ don't commute... | |
| May 9, 2011 at 16:33 | history | answered | Andrey Rekalo | CC BY-SA 3.0 |